Binary Tree Preorder Traversal

Link: http://oj.leetcode.com/problems/binary-tree-preorder-traversal/

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

Recursion version:

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     ArrayList<Integer> nodes = new ArrayList<Integer>();
12 
13     public ArrayList<Integer> preorderTraversal(TreeNode root) {
14 
15         preorderTraversal_recursion(root);
16         return nodes;
17     }
18     public void preorderTraversal_recursion(TreeNode root){
19         if(root==null)
20             return;
21         nodes.add(root.val);
22         preorderTraversal_recursion(root.left);
23         preorderTraversal_recursion(root.right);
24 
25     }
26 }

Iteration version:

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public ArrayList<Integer> preorderTraversal(TreeNode root) {
12     ArrayList<Integer> nodes = new ArrayList<Integer>();
13         if (root == null)
14             return nodes;
15         Stack<TreeNode> stack = new Stack<TreeNode>();
16         // this is used to identify the node pop from stack
17         TreeNode node;
18         stack.push(root);
19         while (!stack.isEmpty()) {
20             node = stack.pop();
21             nodes.add(node.val);
22             if (node.right != null)
23                 stack.push(node.right);
24             if (node.left != null)
25                 stack.push(node.left);
26         }
27         return nodes;
28     }
29 }